\section{Proofs for Section~\ref{sec:fqcount}}\label{app:fq}

We use the notation of Section~\ref{sec:fqcount}, and the following standard facts on
the convergents $p_j/q_j$ of $\beta$ \cite[Chapter~X]{HW08}: $\gcd(p_j,q_j)=1$,
$q_{j+1}=a_{j+1}q_j+q_{j-1}$ and $|q_j\beta-p_j|<1/q_{j+1}$.

\subsection{The slope}
\begin{proof}[Proof of Lemma~\ref{lem:beta}]
(i) We have $q_0=q_1=1$, and $q_j\equiv q_{j-2}\pmod 2$ for $j\ge2$ because $a_j$ is
even; hence every $q_j$ is odd. Let $k\ne0$ and $u$ be integers, $u$ odd, and suppose
that $|2k\beta-u|<1/(4|k|)$. Write $u/(2k)=p/q$ in lowest terms with $q>0$. Since
$\gcd(u,2k)$ is odd, $q$ is even, and $q\le2|k|$. Then
$|\beta-p/q|<1/(8k^2)\le1/(2q^2)$, so $p/q$ is a convergent of $\beta$ by Legendre's
theorem \cite[Theorem~184]{HW08}, and $q$ is odd, a contradiction.

(ii) Let $j\ge1$. Since $q_j$ is odd, there is an integer $r$ with
$|r/q_j-\frac14|=1/(4q_j)$, and since $\gcd(p_j,q_j)=1$, there is
$n\in\{1,\dots,q_j\}$ with $np_j\equiv r\pmod{q_j}$. Put $k=(np_j-r)/q_j$ and
$\eps_j=\beta q_j-p_j$. Then
\[
 \beta n-k-\tfrac14=\bigl(r/q_j-\tfrac14\bigr)+n\eps_j/q_j,
\]
so $|\beta n-k-\frac14|\le1/(4q_j)+1/q_{j+1}\le3/(4n)$, because $n\le q_j$ and
$q_{j+1}\ge a_{j+1}q_j\ge2q_j$. As $j\to\infty$ the bound $1/(4q_j)+1/q_{j+1}$ tends
to zero, whereas $\beta n-\frac14\notin\Z$ for every $n$; hence the integers $n$ obtained
in this way are unbounded.

(iii) Since $\beta$ is irrational, $\eps_j\ne0$, and
$|\eps_j|<1/q_{j+1}\le1/(a_{j+1}q_j)=1/(2q_j^{\,j+2})$.

(iv) By (iii), $|\beta-p_j/q_j|<q_j^{-j-3}$ for all $j\ge1$, and $q_j\to\infty$. If
$\beta$, which is irrational, were algebraic of degree $d$, then $|\beta-p/q|\ge c/q^d$
for all rationals $p/q$ by Liouville's theorem \cite[Theorem~191]{HW08}, which fails
for $j+3>d$.
\end{proof}

\subsection{Geometry of the sheets}
For $0<a<1$ the function $\psi_a$ satisfies $\psi_a(-y)=-\psi_a(y)$,
$\psi_a(y+1)=\psi_a(y)+1$ and $\psi_a(\frac12)=\frac12$, and
\[
 \psi_a'(y)=\frac{1-a^2}{|1-ae(y)|^2}\in\Bigl[\frac{1-a}{1+a},\frac{1+a}{1-a}\Bigr]
\]
is decreasing on $[0,\frac12]$. Hence $\psi_a(y)-y$ vanishes at $0$ and $\frac12$ and
is first increasing and then decreasing on $[0,\frac12]$, so that $\psi_a(y)\ge y$ on
$[0,\frac12]$, $\psi_a(y)\le y$ on $[-\frac12,0]$, and $|\psi_a(y)-y|\le\frac12$ for all
$y$. We write $K_a=(1+a)/(1-a)$ and $g_j(x)=x/\lambda+\psi_{a_j}(\beta x/\lambda-\frac14)$,
so that $A_j=\{x:g_j(x)\in\Z\}$ and $\lambda^{-1}\le g_j'\le\lambda^{-1}(1+\beta K_{a_j})$.

\begin{proof}[Proof of Lemma~\ref{lem:sheets}]
(i) $P$ is admissible of bidegree $(s,s)$. As in the proof of
Lemma~\ref{lem:poisson}(a), $P$ has no zeros in $\{|z|,|w|<1\}$ or in $\{|z|,|w|>1\}$,
and $(Z(x),W(x))$ lies in one of these sets when $x\notin\R$; so all zeros of
$P(Z(x),W(x))$ are real. For real $x$ we have $|W|=1$, hence $1-aW\ne0$, and
$p_a(Z,W)=(1-aW)(1-ZB_a(W))$ vanishes if and only if $ZB_a(W)=1$; for $a=a_j$ this
means $e(g_j(x))=1$, that is, $x\in A_j$. Since $p_b-p_a=(b-a)(z-w)$ and
$p_a(w,w)=1-w^2$, a common zero of $p_a$ and $p_b$, $a\ne b$, satisfies $Z=W=\pm1$.
Then $x/\lambda=u/2$ with $u\in\Z$, and $Z=W$ gives
$(\beta-1)u/2-\frac14\in\Z$, so $2\beta u$ is an odd integer, which is impossible.
Hence the sets $A_j$ are disjoint. At a point of $A_j$, with $a=a_j$,
\begin{equation}\label{eq:sheetderiv}
 \frac{d}{dx}\,p_a(Z(x),W(x))=-(1-aW)\,\frac{d}{dx}e(g_j(x))
 =-2\pi\ii\,(1-aW)\,g_j'(x)\ne0,
\end{equation}
and the other factors of $P$ do not vanish there, so the zeros are simple. Finally,
$g_j(r)-g_j(-r)\le2(1+\beta)r/\lambda+1$ because $|\psi_a(y)-y|\le\frac12$, so
$A_j\cap[-r,r]$ has at most $2(1+\beta)r/\lambda+2$ points.

(ii) Let $|x|<\delta\lambda$ and $t=x/\lambda$. Since $\beta\delta<\frac18$, the number
$y=\beta t-\frac14$ lies in $(-\frac38,-\frac18)$, so
$-\frac12<\psi_{a_j}(y)\le y<-\frac18$ and $g_j(x)=t+\psi_{a_j}(y)\in(-\frac12-\delta,-\frac18+\delta)$,
an interval in $(-1,0)$. Hence $x\notin A$.

(iii) Let $x\in A_j$, $a=a_j$, $t=x/\lambda$, $Z=Z(x)$ and $W=W(x)$. Since
$\frac{d}{dx}p(Z(x),W(x))=\frac{2\pi\ii}{\lambda}D_\beta p(Z,W)$ for every polynomial $p$,
\eqref{eq:sheetderiv} and $p_b(Z,W)=(b-a)(Z-W)$ give
\begin{equation}\label{eq:DbetaP}
 D_\beta P(Z,W)=-(1-aW)\bigl(1+\beta\psi_a'(\beta t-\tfrac14)\bigr)
 \prod_{b\ne a}(b-a)\,(Z-W)^{s-1},
\end{equation}
the product being over the parameters $b\ne a$ of the block. The first factors have
modulus at least $(1-a)\prod_{b\ne a}|b-a|>0$. To bound $\eta=|Z-W|$ from below, note
that $p_a(Z,W)=0$ and $p_a(W,W)=1-W^2$ give $(Z-W)(W-a)=1-W^2$, so that
$|1-W^2|\le2\eta$, and also $|Z^2-W^2|\le2\eta$. Using $|1-e(v)|\ge4\operatorname{dist}(v,\Z)$
we obtain $\operatorname{dist}(2y,\Z)\le\eta/2$ and $\operatorname{dist}(2t-2y,\Z)\le\eta/2$, where
$y=\beta t-\frac14$. Choose integers $u,u'$ with $|2t-u|\le\eta$ and
$|2y-u'|\le\eta/2$. Since $2y=2\beta t-\frac12$,
\[
 |2\beta u-(2u'+1)|\le2|2y-u'|+2\beta|2t-u|\le(1+2\beta)\eta .
\]
If $u\ne0$, Lemma~\ref{lem:beta}(i) gives $(1+2\beta)\eta\ge1/(4|u|)$, and
$|u|\le2|t|+\eta\le2(1+|t|)$; if $u=0$, then $(1+2\beta)\eta\ge1$. In both cases
\begin{equation}\label{eq:ZW}
 |Z-W|\ge\frac{c_\beta}{1+|t|},\qquad c_\beta=\frac1{8(1+2\beta)} .
\end{equation}
Since $\lambda>1$, we have $1+|t|\le1+|x|$, and (iii) follows from \eqref{eq:DbetaP}
and \eqref{eq:ZW}.

(iv) By Lemma~\ref{lem:beta}(ii) there are integers $n$ of arbitrarily large modulus
and integers $k$ with $|\vartheta|\le1/|n|$, where $\vartheta=\beta n-k-\frac14$. Fix such $n$
and $j$, and put $g(t)=t+\psi_{a_j}(\beta t-\frac14)$, so that $g(n)=n+k+\psi_{a_j}(\vartheta)$
with $|\psi_{a_j}(\vartheta)|\le K_{a_j}|\vartheta|$. Since $g'\ge1$, there is $t_j$ with
$g(t_j)=n+k$ and $|t_j-n|\le K_{a_j}/|n|$. The point $x_j=\lambda t_j$ lies in $A_j$, and
$Z(x_j)=e(t_j-n)$ and $W(x_j)=e(\beta(t_j-n)+\vartheta)$ are within $2\pi(1+\beta K_{a_j})/|n|$
of $1$. By \eqref{eq:DbetaP}, $|D_\beta P(Z(x_j),W(x_j))|\le C|n|^{-(s-1)}$, and
$|x_j|$ is comparable to $\lambda|n|$. This proves~(iv); note that the $s$ points
$x_1,\dots,x_s$ all lie within distance $C/|n|$ of $\lambda n$.

(v) Let $x\ne x'$ be points of $A$. If both lie in $A_j$, then $g_j(x)$ and $g_j(x')$
are distinct integers, and $|x-x'|\ge\lambda/(1+\beta K_{a_j})$. Let $x\in A_j$ and
$x'\in A_k$ with $k\ne j$, and let $Z,W,t$ be as in (iii). Then
$e(g_k(x))=ZB_{a_k}(W)$, and $p_{a_k}(Z,W)=(a_k-a_j)(Z-W)$ gives
\[
 |1-e(g_k(x))|=\frac{|p_{a_k}(Z,W)|}{|1-a_kW|}\ge\frac{|a_k-a_j|\,c_\beta}{2(1+|t|)}
\]
by \eqref{eq:ZW}. As $g_k(x')\in\Z$ and $|1-e(v)|\le2\pi\operatorname{dist}(v,\Z)$, we get
$|x-x'|\ge\lambda\operatorname{dist}(g_k(x),\Z)/(1+\beta K_{a_k})\ge\kappa'(1+|x|)^{-1}$.
\end{proof}

\subsection{Choice of the parameters}
\begin{proof}[Proof of Lemma~\ref{lem:fqparams}]
For $a\in(-1,1)$ and $y\in\R$ put
\[
 \psi_a(y)=y+\frac1\pi\arctan\frac{a\sin2\pi y}{1-a\cos2\pi y}.
\]
This is the continuous function with $\psi_a(0)=0$ and $B_a(e(y))=e(\psi_a(y))$, it
is real-analytic in $(a,y)$, $\psi_0(y)=y$, and
$\partial_a\psi_a(y)|_{a=0}=\pi^{-1}\sin2\pi y$. Let $\lambda=cm_r$ and $n\in\Z$. Since
$\partial_t\bigl(t+\psi_a(\beta t-\frac14)\bigr)=1+\beta\psi_a'(\beta t-\frac14)>0$ and
$\psi_a(y)-y$ is bounded, the equation $t+\psi_a(\beta t-\frac14)=n$ has a unique
solution $t_n(a)$, which depends real-analytically on $a\in(-1,1)$ by the implicit
function theorem. For $a\in(0,1)$, the root set of a sheet with parameter $a$ is
$\{x_n(a):n\in\Z\}$, where $x_n(a)=\lambda t_n(a)$.

We have $t_n(0)=(n+\frac14)/(1+\beta)$ and $(1+\beta)t_n'(0)=-\pi^{-1}\sin2\pi y_n$, where
$y_n=\beta t_n(0)-\frac14$. If $2y_n$ were an integer $k$, then
$2\beta(n+\frac14)=(k+\frac12)(1+\beta)$, that is, $\beta(2n-k)=k+\frac12$, which is
impossible. Hence $t_n'(0)\ne0$, and $x_n$ is not constant. Moreover, for $n\ne n'$,
\[
 x_n(0)-x_{n'}(0)=\frac{cm_r(n-n')}{1+\beta}\notin\Gamma_{\Q}:
\]
otherwise $\sqrt2\,m_r(n-n')=(q_1+\beta q_2)(1+\beta)=q_1+(q_1+q_2)\beta+q_2\beta^2$ for
some $q_1,q_2\in\Q$, and since $\beta$ is transcendental (Lemma~\ref{lem:beta}(iv)),
the numbers $1,\beta,\beta^2$ are linearly independent over $\Q(\sqrt2)$, which gives
$q_1=q_2=0$ and then $n=n'$.

We choose $a_{r,1},\dots,a_{r,s_r}$ in turn. When $a_{r,j}$ is chosen, let $\mathcal Y$
be the union of $\{0\}$, the root sets of blocks $1,\dots,r-1$ and the root sets of the
sheets with parameters $a_{r,1},\dots,a_{r,j-1}$; it is countable. We require that
$a=a_{r,j}$ is different from $a_{r,1},\dots,a_{r,j-1}$, that $x_n(a)\notin\rho+\Gamma_{\Q}$
for all $n\in\Z$ and $\rho\in\mathcal Y$, and that $x_n(a)-x_{n'}(a)\notin\Gamma_{\Q}$ for all
$n\ne n'$. For fixed $n$, $\rho$ and $\gamma\in\Gamma_{\Q}$, the function
$a\mapsto x_n(a)-\rho-\gamma$ is real-analytic on $(-1,1)$ and not constant; for fixed
$n\ne n'$ and $\gamma$, the function $a\mapsto x_n(a)-x_{n'}(a)-\gamma$ is real-analytic and
does not vanish at $a=0$. In both cases the zero set in $(0,1)$ is countable, and as
$\Gamma_{\Q}$ is countable, the requirements exclude only countably many $a$. If they hold
at every step, then, by induction, the roots of the new sheets lie outside
$\Gamma_{\Q}$ (take $\rho=0$), differ from all earlier roots by elements outside $\Gamma_{\Q}$,
and in particular are distinct from them because $0\in\Gamma_{\Q}$, and distinct roots of
one sheet differ by elements outside $\Gamma_{\Q}$. Thus blocks $1,\dots,r$ are
independent.
\end{proof}

\subsection{No return to the cone}
\begin{proof}[Proof of Proposition~\ref{prop:noreturn}]
Put $S_1=(C_\beta+F_1)/\lambda$, $S_2=(C_\beta+F_2)/\lambda$ and $\Theta=\lambda^2$, and let
$T\in\Ncal(S_2,S_1)$. Every point $z\in S_1\cup S_2$ has the form $z=(p+\beta q)/\lambda$
with unique integers $p,q$. Since $C_\beta+F_i$ is contained in finitely many
translates of the two quadrants, and $|k+\beta n|=|k|+\beta|n|$ when $k,n$ have the
same sign, there is $K_0$ with $|q|\le K_0(1+|z|)$ for all such $z$. Since $\Theta$ is a
quadratic irrational, Liouville's theorem \cite[Theorem~191]{HW08} gives $c_\Theta>0$
with $|n\Theta-m|\ge c_\Theta/|n|$ for all integers $m$ and $n\ne0$.

Fix $t\in S_1$, write $\lambda t=p_0+\beta q_0$, and choose an integer $k\ge1$ with
$\F T\in H_{-k}$. Put $L=2k+1$ and $M=4k+2$. For $j\ge1$ let $b=q_j$ and
$\eps=\beta q_j-p_j$, so that $0<|\eps|<b^{-j-2}$ by Lemma~\ref{lem:beta}(iii); thus $\eps$ tends to zero faster than every power of $b$ as $j\to\infty$. All
statements below are for sufficiently large $j$, and $C$ denotes constants that do
not depend on $j$.

\emph{Step 1: a test function.} No point of $S_1$ lies within $b^{-4}$ of
$t-\ell\lambda b$ for $1\le\ell\le L$. Indeed, let $z=(p'+\beta q')/\lambda\in S_1$ with
$|z-(t-\ell\lambda b)|\le b^{-4}$. Then $|q'|\le Cb$, and since $\beta b=p_j+\eps$,
\[
 \lambda b\,\bigl(z-t+\ell\lambda b\bigr)
 =\bigl[b(p'-p_0)+p_j(q'-q_0)\bigr]+\eps(q'-q_0)+\ell\Theta b^2 .
\]
The bracket is an integer, so the sum of the first and the last term has modulus at
least $c_\Theta/(Lb^2)$, while $|\eps(q'-q_0)|\le C|\eps|b<c_\Theta/(2Lb^2)$. Hence
$\lambda b^{-3}\ge c_\Theta/(2Lb^2)$, which is false for large $b$.

Let $\phi\in C_c^\infty((-1,1))$ with $\phi(0)=1$, $f_b(x)=\phi(b^4(x-t))$ and
\[
 \eta_b(x)=\sum_{\ell=0}^L(-1)^\ell\binom L\ell f_b(x+\ell\lambda b).
\]
Since $S_1$ is locally finite, the interval $(t-b^{-4},t+b^{-4})$ contains no point
of $S_1$ other than $t$, and by Step~1 the other bumps contain no point of $S_1$.
Hence $T(\eta_b)=T(\{t\})$. With $h_b=\F^{-1}f_b$ and $g_b(\xi)=1-e(-\lambda b\xi)$ we
have $\F^{-1}\eta_b=g_b^Lh_b$, and since $h_b(\xi)=b^{-4}e(t\xi)(\F^{-1}\phi)(\xi/b^4)$,
\begin{equation}\label{eq:hb}
 |h_b^{(r)}(\xi)|\le C_{r,N}\,b^{-4}(1+|\xi|/b^4)^{-N}\qquad(r,N\ge0).
\end{equation}

\emph{Step 2: the spectral grid.} Put $R=b^8|\eps|^{-1/(2M)}$ and $w=K|\eps|R/b$ with
$K=12K_0/\lambda$. If $\xi=(p+\beta q)/\lambda\in S_2$ and $|\xi|\le2R$, then
$|q|\le K_0(1+2R)\le3K_0R$ and $\lambda b\xi=(bp+p_jq)+\eps q$, so
\[
 \operatorname{dist}\bigl(\xi,(\lambda b)^{-1}\Z\bigr)\le|q||\eps|/(\lambda b)\le w/4 .
\]
Moreover $bw=Kb^8|\eps|^{1-1/(2M)}\to0$ and $R/b^4=b^4|\eps|^{-1/(2M)}\to\infty$. Let
$\chi_1,\chi_2\in C_c^\infty((-2,2))$ be equal to $1$ on $[-1,1]$, and put
$\chi_R(\xi)=\chi_1(\xi/R)$ and $\chi_{\mathrm{gr}}(\xi)=\sum_{v\in\Z}\chi_2((\xi-v/(\lambda b))/w)$.
As $4w<1/(\lambda b)$, the windows $|\xi-v/(\lambda b)|<2w$ are disjoint,
$\chi_{\mathrm{gr}}$ equals $1$ within distance $w$ of the grid, and
$|\chi_{\mathrm{gr}}^{(r)}|\le C_rw^{-r}$. Let
\[
 K_b=K_{\mathrm{in}}+K_{\mathrm{out}},\qquad
 K_{\mathrm{in}}=\chi_R\,\chi_{\mathrm{gr}}\,g_b^Lh_b,\qquad
 K_{\mathrm{out}}=(1-\chi_R)\,g_b^Lh_b .
\]
Then $K_b\in\Sc$, and $K_b-g_b^Lh_b=\chi_R(\chi_{\mathrm{gr}}-1)g_b^Lh_b$ belongs to
$C_c^\infty(\R)$ and vanishes at every point of $S_2$. Since $\F T$ is locally atomic on
$S_2$,
\begin{equation}\label{eq:noreturnid}
 T(\{t\})=T(\eta_b)=\F T(\F^{-1}\eta_b)=\F T(K_b).
\end{equation}

\emph{Step 3: estimates.} By Section~\ref{sec:prelim} it suffices to show that
$\norm{\xi^u\partial^rK_{\mathrm{in}}}_{L^2}$ and $\norm{\xi^u\partial^rK_{\mathrm{out}}}_{L^2}$
tend to zero for $u+r\le2k$; then $\norm{K_b}_k\to0$, and \eqref{eq:noreturnid} and
$|\F T(K_b)|\le\norm{\F T}_{-k}\norm{K_b}_k$ give $T(\{t\})=0$.

The function $K_{\mathrm{in}}$ is supported in the union of the $O(bR)$ windows that
meet $[-2R,2R]$, a set of measure $O(bRw)$. On a window, $|g_b|\le4\pi\lambda bw$ and,
for $v\le2k<L$, $|(g_b^L)^{(v)}|\le C(bw)^{L-v}b^v=C(bw)^Lw^{-v}$. A derivative that
falls on $\chi_R$, $\chi_{\mathrm{gr}}$ or $h_b$ costs at most $Cw^{-1}$, as $w<1<R$, and
$|h_b|\le Cb^{-4}$. Hence, using $bw=K|\eps|R$,
\[
 \norm{\xi^u\partial^rK_{\mathrm{in}}}_{L^2}\le CR^u(bw)^Lw^{-r}b^{-4}(bRw)^{1/2}
 \le C|\eps|^{L-r+\frac12}\,b^{r-4}R^{u+L-r+1}.
\]
Here $u+L-r+1\le M$, so substituting $R=b^8|\eps|^{-1/(2M)}$ lowers the exponent of
$|\eps|$ by at most $\frac12$, and we obtain the bound $C|\eps|^{L-r}b^{2k-4+8M}\le Cb^{2k-4+8M}|\eps|$, which tends to zero.

On the support of $K_{\mathrm{out}}$ we have $|\xi|\ge R$, and there
$|(g_b^L)^{(v)}|\le Cb^v$. By \eqref{eq:hb} with $N=2k+2$,
$|\xi^u\partial^rK_{\mathrm{out}}(\xi)|\le Cb^{r-4}|\xi|^u(|\xi|/b^4)^{-N}$, and therefore
\[
 \norm{\xi^u\partial^rK_{\mathrm{out}}}_{L^2}\le Cb^{r-4+4N}R^{u-N+\frac12}
 \le Cb^{8k-2}(R/b^4)^{-3/2}=Cb^{8k-8}|\eps|^{3/(4M)},
\]
which tends to zero. Thus $T(\{t\})=0$ for every $t\in S_1$, and $T=0$.
\end{proof}

\subsection{Sparse splitting}
Let $X=\mathbb P^1\times\mathbb P^1\supset\C^2$. For an effective divisor $D$ on $X$ we
write $\mathcal O(D)$ for the sheaf of rational functions $h$ with $\operatorname{div}h+D\ge0$
(together with $0$). The group $G_i=\mu_{m_i}\times\mu_{m_i}$ acts on $X$ by
$(z,w)\mapsto(\zeta z,\zeta'w)$, and on rational functions by $gF=F\circ g^{-1}$; the
polar divisor of $gF$ is the image under $g$ of that of $F$. The groups $G_i$
commute. Let $D_i$ be the closure in $X$ of the curve $\{Q_i=0\}\subset\C^2$, and
$D=\sum_iD_i$. Since $Q_i$ has nonzero coefficients at $1$, $z^{d_i}$, $w^{d_i}$ and
$z^{d_i}w^{d_i}$, where $d_i=n_is_i$, the divisor of the bihomogenization of $Q_i$ has no
component at infinity; as $Q_i$ is square-free (Step~1 below), this divisor is
$D_i$.

\begin{proof}[Proof of Lemma~\ref{lem:splitting}]
\emph{Step 1: the components.} Let $a\in(0,1)$, $|\rho|=|\sigma|=1$ and $n\ge1$. The
polynomial $p_a(\rho z^n,\sigma w^n)=(1-a\sigma w^n)-\rho z^n(\sigma w^n-a)$ is irreducible: over
$\C(w)$ it is a multiple of $z^n-h(w)$ with
$h=(1-a\sigma w^n)/(\rho(\sigma w^n-a))$, and $h$ has a simple zero at every root of
$1-a\sigma w^n$, which is not a root of $\sigma w^n-a$ because $a^2\ne1$; so $z^n-h$ is
Eisenstein at such a root and irreducible over $\C(w)$, and since the coefficients
$1-a\sigma w^n$ and $-\rho(\sigma w^n-a)$ are coprime, the polynomial is primitive in
$\C[w][z]$ and irreducible by Gauss's lemma. Its coefficients at $1$, $z^n$ and $w^n$
are $1$, $a\rho$ and $-a\sigma$, so two such polynomials with the same $n$ are
associates only if they have the same $(a,\rho,\sigma)$; polynomials with different $n$
have different bidegrees. As $n_j$ is prime to $m_j$, the map $\zeta\mapsto\zeta^{n_j}$ is a
bijection of $\mu_{m_j}$. Hence $Q$ is square-free, $D$ is reduced, and a
nontrivial $g\in G_j$ maps every component of $D_j$ to a curve that is not a
component of $D$, nor of $g'D_i$ for any $i\ne j$ and $g'\in G_i$. Finally $D_k$ is
$G_j$-invariant for $k\ne j$, because $m_j$ divides $n_k$.

\emph{Step 2: no base points.} For every $i$ and every point $\mathrm P\in X$ there is
$g\in G_i$ with $\mathrm P\notin gD_i$. Let $\mathrm P=(z,w)$ with $z,w\in\C\setminus\{0\}$,
and put $U=z^{n_i}$ and $V=-\ii w^{n_i}$. If the points $(\zeta z,w)$, $\zeta\in\mu_{m_i}$, all
lay on $D_i$, then for every $\varrho\in\mu_{m_i}$ there would be a parameter $a$ of block
$i$ with $\varrho U=(1-aV)/(V-a)$. If $V=\pm1$, the right-hand side equals $V$ for every
$a$, which is impossible since $m_i\ge3$. Otherwise the right-hand side is injective
in $a$, so at least three parameters $a$ satisfy
$|1-aV|^2-|U|^2|V-a|^2=0$; this real quadratic polynomial in $a$ then vanishes
identically, which gives $|U|=|V|=1$. But for $|V|=1$,
\[
 \operatorname{Im}\frac{1-aV}{V-a}=\frac{(a^2-1)\operatorname{Im}V}{|V-a|^2},
\]
so all the points $\varrho U$ would lie in one open half-plane bounded by $\R$, whereas
their sum is zero. At the remaining points one of the coordinates is $0$ or
$\infty$. On the lines $z=0$, $z=\infty$, $w=0$ and $w=\infty$ the factors of block $i$
are, up to nonzero constants, $1-aV$, $a-V$, $1+aU$ and $a+U$, which vanish only
for $V>1$, $0<V<1$, $U<-1$ and $-1<U<0$, respectively. At a point of one of these
lines that is not a corner, the relevant coordinate $V$ or $U$ is nonzero, and a
rotation of $w$ or of $z$ by an element of $\mu_{m_i}$ moves it off the corresponding
half-line, since $m_i\ge3$. The four corners do not lie on $D_i$, since the
corresponding coefficients of $Q_i$ are nonzero.

\emph{Step 3: mixed differences.} Put $F=R/Q$. On $\C^2$ the poles of $F$ lie on
$\{Q=0\}$ and are simple, and at infinity $F$ is the quotient of the
bihomogenizations of $R$ and $Q$ in bidegree $(S,S)$; hence $F\in H^0(X,\mathcal O(D))$.
An element $(\zeta,\zeta')\in G_j$ multiplies the coefficient of $z^kw^n$ in the Taylor
series of $F$ at the origin by $\zeta^{-k}\zeta'^{-n}$, and a monomial with
$(k,n)\in n_l\Z^2$ is invariant under $G_j$ for every $j\ne l$. Hence for $i\ne j$,
$g_i\in G_i$ and $g_j\in G_j$, every monomial of the Taylor series is annihilated by
$g_i-1$ or by $g_j-1$, and
\begin{equation}\label{eq:mixed}
 (g_i-1)(g_j-1)F=0 .
\end{equation}

\emph{Step 4: the poles of a difference.} Let $g\in G_i$ and $\Delta=gF-F$. By
\eqref{eq:mixed}, $\Delta$ is $G_j$-invariant for every $j\ne i$. Its poles are simple and
lie on $D\cup gD=D_i\cup gD_i\cup\bigcup_{j\ne i}D_j$. If $\Delta$ had a pole along a
component $Y$ of some $D_j$, $j\ne i$, it would also have a pole along $g'Y$ for every
$g'\in G_j$, which is impossible for $g'\ne1$ by Step~1. Hence
\begin{equation}\label{eq:diffpoles}
 gF-F\in H^0\bigl(X,\mathcal O(D_i+gD_i)\bigr)\quad(g\ne1),\qquad
 gF-F\in H^0\bigl(X,\mathcal O(D_i)\bigr)\quad(g=1).
\end{equation}

\emph{Step 5: splitting off one block.} Fix $i$ and let $D'=\sum_{j\ne i}D_j$, which is
$G_i$-invariant. Given $\mathrm P\in X$, choose $g\in G_i$ with $\mathrm P\notin gD_i$
(Step~2). On the open set $X\setminus gD_i$ we have $F=gF-(gF-F)$, where the poles of
$gF$ lie on $gD=gD_i+D'$, so that $gF$ is a section of $\mathcal O(D')$ there, and
$-(gF-F)$ is a section of $\mathcal O(D_i)$ there by \eqref{eq:diffpoles}. Thus $F$ is a
global section of the subsheaf $\mathcal O(D_i)+\mathcal O(D')$ of the constant sheaf of
rational functions. Since $D_i$ and $D'$ have no common component and $X$ is
smooth, $\mathcal O(D_i)\cap\mathcal O(D')=\mathcal O_X$ \cite[Chapter~II,
Proposition~6.3A]{Har77}, so the sequence
\[
 0\to\mathcal O_X\to\mathcal O(D_i)\oplus\mathcal O(D')\to\mathcal O(D_i)+\mathcal O(D')\to0,
\]
with maps $h\mapsto(h,-h)$ and $(a,b)\mapsto a+b$, is exact. As
$H^1(X,\mathcal O_X)=0$ \cite[Chapter~III, Exercise~5.6(a)]{Har77}, the long exact
sequence of cohomology gives $F=F_i+F^{(i)}$ with $F_i\in H^0(X,\mathcal O(D_i))$ and
$F^{(i)}\in H^0(X,\mathcal O(D'))$.

\emph{Step 6: the remainder is constant.} Do this for every $i$ and put
$G=F-\sum_iF_i$. Along a component of $D_i$ the functions $F-F_i=F^{(i)}$ and $F_j$,
$j\ne i$, have no poles. So $G$ has no poles on $X$, and it is constant.

\emph{Step 7: invariance.} Let $j\ne i$ and $g\in G_j$. Then
$(g-1)F_i=(g-1)F-\sum_{k\ne i}(g-1)F_k$. The left-hand side has poles only on $D_i$,
which is $G_j$-invariant. On the right-hand side, $(g-1)F$ has poles only on
$D_j\cup gD_j$ by \eqref{eq:diffpoles}, $(g-1)F_k$ has poles only on $D_k$ for
$k\ne j$, and $(g-1)F_j$ only on $D_j\cup gD_j$. Hence $(g-1)F_i$ is a constant
$c(g)$. Since constants are invariant, $c(gg')=c(g)+c(g')$, so $c$ is a homomorphism
from the finite group $G_j$ to $(\C,+)$, and $c=0$. Thus $F_i$ is invariant under
$H_i=\prod_{j\ne i}G_j$.

\emph{Step 8: descent.} Since $F_i$ has at most simple poles along $D_i$ and $Q_i$ is
square-free, $J_i=F_iQ_i$ has no poles on $\C^2$, so it is a polynomial. For all but
finitely many $w_0$, the function $F_i(\cdot,w_0)$ is regular at $z=\infty$ and
$Q_i(\cdot,w_0)$ has degree $d_i$; hence $J_i$ has degree at most $d_i$ in $z$, and
likewise in $w$. As $F_i$ and $Q_i$ are $H_i$-invariant, so is $J_i$, and since
$n_i=\prod_{j\ne i}m_j$, every monomial $z^kw^n$ of $J_i$ has $n_i\mid k$ and $n_i\mid n$.
Thus $J_i=R_i'(z^{n_i},-\ii w^{n_i})$ for a polynomial $R_i'(Z,W)$ of bidegree at most
$(s_i,s_i)$. Let $\tau_i$ be the quotient of the coefficients of $Z^{s_i}W^{s_i}$ in $R_i'$
and in $P_i$, and $R_i=R_i'-\tau_iP_i\in V_{s_i}$; then $F_i=R_i(z^{n_i},-\ii w^{n_i})/Q_i+\tau_i$,
and
\[
 \frac RQ=\sum_{i\le r}\frac{R_i(z^{n_i},-\ii w^{n_i})}{Q_i(z,w)}+G',\qquad
 G'=G+\sum_i\tau_i .
\]
All terms are regular at $(\infty,\infty)$, and their values there are the quotients
of the top coefficients of numerator and denominator. These vanish for $R/Q$,
because $R\in V_S$, and for every summand, by the choice of $R_i$. Hence $G'=0$.
\end{proof}